function - 结合 (a -> Maybe a) 和 (a -> a) 函数

Question

我有两个功能：

f :: a -> Maybe a
g :: a -> a

我想创建这样的功能：

h :: a -> Maybe a

h x
| isJust(f x) = Just (g $ fromJust(f x))
| otherwise   = Nothing

我怎样才能以更优雅的方式做到这一点？

Answer 1

由于您已使用dot-operator标记了此问题：

h :: a -> Maybe a
h = fmap g . f

解释：

f            ::                          a -> Maybe a
g            ::        a ->       a
fmap g       ::  Maybe a -> Maybe a
(.)          :: (Maybe a -> Maybe a) -> (a -> Maybe a) -> (a -> Maybe a)
(.) (fmap g) ::                         (a -> Maybe a) -> (a -> Maybe a)
fmap g . f   ::                                           (a -> Maybe a)
h            ::                                            a -> Maybe a

请注意，(.)' 和fmap g' 类型实际上更通用：

(.) :: (b -> c) -> (a -> b) -> (a -> c)
-- b in this case is Maybe a
-- c in this case is Maybe a

fmap g :: Functor f => f a -> f a
-- f in this case is Maybe

但是，您也可以对以下结果进行模式匹配f：

h x = 
  case f x of
    Just k -> Just (g k)
    _      -> Nothing

请注意，您的原始示例甚至无法编译，因为g' 的返回类型不正确。

Answer 2

有

fmap2 :: (Functor g, Functor f) => (a -> b) -> g (f a) -> g (f b)
fmap2 = fmap . fmap

这是一个有趣的方式：

h :: a -> Maybe a
h = fmap2 g f

fmap2 g f ~> fmap (fmap g) f ~> fmap g . f ~> \x -> fmap g (f x)

此处使用Functor ((->) r)实例：fmap可用于代替(.).

Answer 3

为什么不简单地这样做：

h :: a -> Maybe a
h x = fmap g (f x)

或运营商版本：

h :: a -> Maybe a
h x = g <$> f x